Question

At a certain pizza parlor, 43% of the customers order a pizza containing onions, 33% of the customers order a pizza containing sausage, and 67% order a pizza containing onions or sausage (or both). Find the probability that a customer chosen at random will order a pizza containing both onions and sausage.

Answer 1

To find the probability of a pizza containing both onions and sausage, we used the principle of inclusion-exclusion, which resulted in a calculation showing that the probability is 9%.

Step-by-step explanation:

To find the probability that a customer chosen at random will order a pizza containing both onions and sausage, we can use the principle of inclusion-exclusion in probability. According to the data provided, 43% of the customers order pizza with onions, 33% with sausage, and 67% order pizza containing either onions or sausage (or both). The principle of inclusion-exclusion states that the probability of either event A (onions) or event B (sausage) occurring is the sum of the probabilities of each event occurring separately minus the probability of both events occurring together. This is given by the formula P(A OR B) = P(A) + P(B) - P(A AND B). We use it to solve for P(A AND B) (the probability of both toppings), which is the unknown we are looking for.

According to the formula:

• P(Onions) = P(A) = 0.43
• P(Sausage) = P(B) = 0.33
• P(Onions OR Sausage) = P(A OR B) = 0.67

P(A AND B) = P(A) + P(B) - P(A OR B)

P(A AND B) = 0.43 + 0.33 - 0.67 = 0.09 or 9%

Therefore, the probability that a customer chosen at random will order a pizza containing both onions and sausage is 9%.

Answer 2

Answer:

9%

Step-by-step explanation:

The percentage that a customer request for a pizza that contains onion = 43%

The percentage that a customer request for a pizza that contains sausage = 33%

The percentage that a customer request for a pizza that contains onion and/or sausage = 67%

We can represent this using probability notations

P(Onions) = 43% = 0.43

P(Sausage) = 33% = 0.33

P(Onions and/or Sausage) = 67% = 0.67

P(Onions and/or Sausage) can be translated to customers that request for pizza containing "Onions only" , "Sausage only", "Onions and Sausage"

Mathematical, we represent the above statement:

P(Onions) + P(Sausages) - P(Onions and Sausage)

So, we have

P(Onions and/or Sausage) = P(Onions) + P(Sausages) - P(Onions and Sausage)

0.67 = 0.43 + 0.33 - P(Onions and Sausage)

0.67= 0.76 - P(Onions and Sausage)

P(Onions and Sausage) = 0.76 - 0.67

P(Onions and Sausage) = 0.9

So, the probability of a customer making an order of pizza that contains onions and sausage is 0.9 or 9%